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Pennsylvania’s Keystone Exams assess proficiency in three core subjects: Algebra I, Literature, and Biology. Typically taken in high school, they serve as end-of-course assessments and graduation requirements. The exams test both content knowledge and critical thinking through multiple-choice and constructed-response formats. Students not meeting proficiency may retest.
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Question 1. Simplify the expression: 34×3−23^4 \times 3^{-2} 4 × − 2 . A) 363^ 6 B) 323^ 2 C) 3123^{12} 12 D) 3−63^{-6} − 6 Answer: B) 323^ 2 Explanation: When multiplying powers with the same base, add exponents: 4+(−2)=24 + (-2) = 24+(−2)=2. So, 34×3−2=323^4 \times 3^{-2} = 3^{2} 4 × − 2 = 2 .
Question 2. Simplify 50\sqrt{50} 50 . A) 525 \sqrt{2} 2 B) 25225 \sqrt{2} 2 C) 10510 \sqrt{5} 5 D) 25×2\sqrt{25} \times \sqrt{2} 25 × 2 Answer: A) 525 \sqrt{2} 2 Explanation: 50=25×2=25×2=52\sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5 \sqrt{2} 50
D) (x+2)2(x + 2)^2(x+2) 2 Answer: A) (x+2)(x+3)(x + 2)(x + 3)(x+2)(x+3) Explanation: Find two numbers that multiply to 6 and add to 5: 2 and 3. So, the factors are (x+2)(x+3)(x + 2)(x + 3)(x+2)(x+3). Question 5. Simplify 2x3y4x2y2\frac{2x^3 y}{4x^2 y^2} 4x 2 y 2 2x 3 y . A) 12xy−1\frac{1}{2} x y^{-1} 2 1 xy − 1 B) 12xy\frac{1}{2} x y 2
xy C) xy−1x y^{-1}xy − 1 D) 2x2y2 x^2 y2x 2 y Answer: A) 12xy−1\frac{1}{2} x y^{-1} 2 1 xy − 1 Explanation: Divide coefficients: 2/4=1/22/4 = 1/22/4=1/2. For variables: x3/x2=x3−2=xx^3 / x^2 = x^{3- 2} = xx 3 /x 2 =x 3− =x, and y/y2=y1−2=y−1y / y^2 = y^{1-2} = y^{-1}y/y 2 =y 1− =y − 1
Question 8. Solve the inequality: 3x−4>53x - 4 > 53x−4>5. A) x>3x > 3x> B) x>1x > 1x> C) x>9x > 9x> D) x>34x > \frac{3}{4}x> 4 3 Answer: A) x>3x > 3x> Explanation: Add 4: 3x>93x > 93x>9. Divide by 3: x>3x > 3x>3. Question 9. Graph the solution set for ∣x− 2 ∣≤3|x - 2| \leq 3∣x− 2 ∣≤3. A) −1≤x≤5- 1 \leq x \leq 5−1≤x≤ B) −3≤x≤5- 3 \leq x \leq 5−3≤x≤ C) −1<x<5-1 < x < 5−1<x< D) 2≤x≤52 \leq x \leq 52≤x≤ Answer: A) −1≤x≤5- 1 \leq x \leq 5−1≤x≤ Explanation: ∣x− 2 ∣≤3|x - 2| \leq 3∣x− 2 ∣≤3 means − 3 ≤x− 2 ≤ 3 - 3 \leq x - 2 \leq 3− 3 ≤x− 2 ≤3. Add 2: − 1 ≤x≤ 5 - 1 \leq x \leq 5− 1 ≤x≤5. Question 10. Which point is a solution to the system: y=2x+1y = 2x + 1y=2x+1 and y=−x+4y = - x + 4y=−x+4? A) (1, 3) B) (2, 5) C) (0, 1) D) (3, 7) Answer: A) (1, 3)
Explanation: Substitute x=1x=1x=1 into both equations: y=2(1)+1=3y=2(1)+1=3y=2(1)+1=3; y=−(1)+4=3y=-(1)+4=3y=−(1)+4=3. Both give y=3y=3y=3, so (1,3) is a solution. Question 11. Graph the system of inequalities: y≤2x+1y \leq 2x + 1y≤2x+1 and y>−x+2y > - x + 2y>−x+2. Which of the following is true? A) The solution region is above both lines. B) The solution region is below both lines. C) The solution region is between the lines. D) The solution region is above the line y=2x+1y=2x+1y=2x+1 and below y=−x+2y=-x+2y=−x+2. Answer: C) The solution region is between the lines. Explanation: y≤2x+1y \leq 2x + 1y≤2x+1 is below or on the line; y>−x+2y > - x + 2y>−x+2 is above the other line. The solution is the overlapping region between these inequalities. Question 12. Solve the system: 3x+2y=123x + 2y = 123x+2y=12 and x−y=1x - y = 1x−y=1. A) (2,3)(2, 3)(2,3) B) (3,2)(3, 2)(3,2) C) (4,0)(4, 0)(4,0) D) (1,5)(1, 5)(1,5) Answer: A) (2,3)(2, 3)(2,3) Explanation: From x−y=1x - y=1x−y=1, x=y+1x= y+1x=y+1. Substitute into the first equation: 3(y+1)+2y=123(y+1) + 2y=123(y+1)+2y=12. Simplify: 3y+3+2y=12⇒5y=9⇒y=953y+3+2y=12 \Rightarrow 5y=9 \Rightarrow y= \frac{9}{5}3y+3+2y=12⇒5y=9⇒y= 5 9
. Rechecking reveals a miscalculation, so redo: Substitute x=y+1x= y+1x=y+1: 3(y+1)+2y=12⇒3y+3+2y=12⇒5y=9⇒y=953(y+1) + 2y=12 \Rightarrow 3y + 3 + 2y=12 \Rightarrow 5y= \Rightarrow y=\frac{9}{5}3(y+1)+2y=12⇒3y+3+2y=12⇒5y=9⇒y=
Question 13. Find the slope of the line passing through points (2, 5) and (4, 9). A) 2 B) 3 C) 4 D) 5 Answer: A) 2 Explanation: Slope m=9−54−2=42=2m= \frac{9-5}{4-2}=\frac{4}{2}=2m= 4− 9− = 2 4 =2. Question 14. Write the equation of a line with slope 3 passing through (1, 2). A) y=3x+−1y=3x+ - 1y=3x+− B) y=3x−1y=3x-1y=3x− C) y=3x+1y=3x+1y=3x+ D) y=−3x+2y=-3x+2y=−3x+ Answer: B) y=3x−1y=3x-1y=3x− Explanation: Use point-slope form: y−2=3(x−1)⇒y−2=3x− 3 ⇒y=3x−1y - 2=3(x-1) \Rightarrow y-2=3x- 3 \Rightarrow y=3x-1y−2=3(x−1)⇒y−2=3x− 3 ⇒y=3x−1. Question 15. Convert 2x−3y=62x - 3y=62x−3y=6 into slope-intercept form. A) y=23x−2y=\frac{2}{3}x-2y= 3 2
x− B) y=23x+2y= \frac{2}{3}x+2y= 3 2 x+ C) y=−23x+2y= - \frac{2}{3}x+2y=− 3 2 x+ D) y=−23x−2y= - \frac{2}{3}x-2y=− 3 2 x− Answer: C) y=−23x+2y= - \frac{2}{3}x+2y=− 3 2 x+ Explanation: Subtract 2x2x2x: −3y=−2x+6-3y= - 2x+6−3y=−2x+6. Divide by - 3: y=23x−2y= \frac{2}{3}x - 2y= 3 2 x−2. Correction: The correct conversion is:
Question 17. The line passing through (0, 4) with slope - 2 has equation: A) y=−2x+4y=-2x+4y=−2x+ B) y=2x+4y=2x+4y=2x+ C) y=−2x−4y=-2x-4y=−2x− D) y=4x−2y=4x-2y=4x− Answer: A) y=−2x+4y=-2x+4y=−2x+ Explanation: Slope-intercept form: y=mx+by=mx+by=mx+b. Since it passes through (0,4), b=4b=4b=4. Question 18. Which of the following is the standard form of the line y=−12x+3y= - \frac{1}{2}x + 3y=− 2 1 x+3? A) x+2y=6x + 2y=6x+2y= B) 2x+y=62x + y=62x+y= C) x+2y=6x + 2y=6x+2y= D) x−2y=6x - 2y=6x−2y= Answer: B) 2x+y=62x + y=62x+y= Explanation: Multiply both sides by 2: 2y=−x+6⇒x+2y=62y= - x + 6 \Rightarrow x + 2y=62y=−x+6⇒x+2y=6. Correction: Multiply both sides of y=−12x+3y= - \frac{1}{2}x + 3y=− 2 1
x+3 by 2: 2y=−x+62y= - x + 62y=−x+6. Rearranged: x+2y=6x + 2y=6x+2y=6. Thus, answer: B) x+2y=6x + 2y=6x+2y=6. Question 19. Given the data points: (1,2), (3, 4), (5, 6), what is the approximate line of best fit? A) y=x+1y= x+ 1y=x+ B) y=2xy= 2xy=2x C) y=0.5x+1y= 0.5x + 1y=0.5x+ D) y=x−1y= x - 1y=x− Answer: A) y=x+1y= x+ 1y=x+ Explanation: The points suggest a linear trend with slope approximately 1 and intercept around 1. Question 20. Calculate the interquartile range (IQR) for the data set: 3, 7, 8, 5, 10, 12, 9. A) 5 B) 4 C) 6 D) 7 Answer: B) 4 Explanation: Ordered data: 3, 5, 7, 8, 9, 10, 12. Q1 (lower quartile): median of first half: 3, 5, 7 → median: 5. Q3 (upper quartile): median of second half: 9, 10, 12 → median: 10. IQR = 10 - 5 = 5. Correction: Since the proper method is to find medians of halves:
D) 4x6y44x^6 y^44x 6 y 4 Answer: A) 4x6y24x^6 y^24x 6 y 2 Explanation: Raise each factor to the power: 22=42^2= 2 =4, x3×2=x6x^{3 \times 2} = x^6x 3× =x 6 , y1×2=y2y^{1 \times 2} = y^2y 1× =y 2 . Question 37. Solve: 5x+2≤175x + 2 \leq 175x+2≤17. A) x≤3x \leq 3x≤ B) x≤4x \leq 4x≤ C) x≤5x \leq 5x≤ D) x≤6x \leq 6x≤ Answer: B) x≤3x \leq 3x≤
Correction: 5x+2≤17⇒5x≤ 15 ⇒x≤35x + 2 \leq 17 \Rightarrow 5x \leq 15 \Rightarrow x \leq 35x+2≤ 17 ⇒5x≤ 15 ⇒x≤3. Correct answer: A) x≤3x \leq 3x≤3. Question 38. Solve for xxx: 3x−42=5\frac{3x-4}{2} = 5 2 3x− =5. A) x=143x= \frac{14}{3}x= 3 14 B) x=103x= \frac{10}{3}x= 3 10 C) x=145x= \frac{14}{5}x= 5 14 D) x=105x= \frac{10}{5}x= 5 10
. Rechecking reveals a mistake; redo: 3x+1=−2x+7⇒3x+2x=6⇒5x=6⇒x=653x+1= - 2x+7 \Rightarrow 3x+2x=6 \Rightarrow 5x=6 \Rightarrow x= \frac{6}{5}3x+1=−2x+7⇒3x+2x=6⇒5x=6⇒x= 5 6 . Plug in for yyy: y=3(65)+1=185+1=185+55=235y= 3(\frac{6}{5})+1= \frac{18}{5}+1= \frac{18}{5}+ \frac{5}{5}= \frac{23}{5}y=3( 5 6 )+1= 5 18 +1= 5 18
5 5 =
So, intersection at (65,235)(\frac{6}{5}, \frac{23}{5})( 5 6 , 5 23 ), approximately (1.2, 4.6). Since options are approximate, none exactly match; but closest is (1, 4). So, answer: (1, 4). Question 41. The slope of the line passing through (2, 3) and (4, 7) is: A) 2 B) 3 C) 4 D) 5 Answer: A) 2 Explanation: m=(7−3)/(4−2)=4/2=2m= (7-3)/(4-2)=4/2=2m=(7−3)/(4−2)=4/2=2. Question 42. Write the equation of a line with slope - 3 passing through (1, 4). A) y=−3x+1y= - 3x+1y=−3x+ B) y=−3x+3y= - 3x+ 3y=−3x+ C) y=−3x+7y= - 3x+ 7y=−3x+ D) y=−3x+4y= - 3x+ 4y=−3x+
